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Proof step 5 from proof And-forall (previous step)

((aP0 ∧_ s))(H), ∀ a, H. a ∈ s ∧ aP0(a)(H) → aP(a)(H), ∀ s0, aP0, H, aP. # s0 < # s → ((aP0 ∧_ s0))(H) ∧ (∀ a, H. a ∈ s0 ∧ aP0(a)(H) → aP(a)(H)) → ((aP ∧_ s0))(H) ⊦ ((aP ∧_ s))(H)

User applied rule cut formula

And obtained the following new goals

Proof step 6:
((aP0 ∧_ s))(H), ∀ a, H. a ∈ s ∧ aP0(a)(H) → aP(a)(H), ∀ s0, aP0, H, aP. # s0 < # s → ((aP0 ∧_ s0))(H) ∧ (∀ a, H. a ∈ s0 ∧ aP0(a)(H) → aP(a)(H)) → ((aP ∧_ s0))(H) ⊦ s = ∅, ((aP ∧_ s))(H)

Rule argument

fma: s = ∅

Additional information

Used simplifier rules: And-emp: ⊦ aP ∧_ ∅ = true;% ⊦ true(H); (s) Used formulas: s = ∅